Find the area bounded by the loop of the Folium

Find the area bounded by the loop of the Folium of Descartes: $x = \frac{3at}{1 + t^3}$, $y = \frac{3at^2}{1 + t^3}$.

We will apply the parametric area formula over the full loop and evaluate the resulting rational integral by substitution.

Step 1: Understand the Folium of Descartes

In Cartesian form the Folium is $x^3+y^3=3axy$. In the given parametric form, $t=0$ corresponds to the origin, both $x,y\to0$ as $t\to\infty$, and the loop is traced for $t\in(0,\infty)$.

Step 2: Compute the derivative $\frac{dx}{dt}$

\[\frac{dx}{dt} = \frac{3a(1+t^3) - 3at\cdot3t^2}{(1+t^3)^2} = \frac{3a(1-2t^3)}{(1+t^3)^2}\]

Step 3: Set up the area integral

Using $A=\int_0^\infty y\,\frac{dx}{dt}\,dt$:

\[A = \int_0^{\infty} \frac{3at^2}{1+t^3}\cdot\frac{3a(1-2t^3)}{(1+t^3)^2}\,dt = 9a^2\int_0^{\infty} \frac{t^2(1-2t^3)}{(1+t^3)^3}\,dt\]

Step 4: Split and evaluate using $u=1+t^3$

\[A = 9a^2\left[\int_0^{\infty} \frac{t^2}{(1+t^3)^3}\,dt - 2\int_0^{\infty} \frac{t^5}{(1+t^3)^3}\,dt\right]\]

With $u=1+t^3$, $du=3t^2\,dt$:

\[\int_0^{\infty} \frac{t^2}{(1+t^3)^3}\,dt = \frac13\int_1^{\infty}\frac{du}{u^3} = \frac13\left[-\frac{1}{2u^2}\right]_1^{\infty} = \frac16\]

Writing $t^5=t^3\cdot t^2=(u-1)t^2$:

\[\int_0^{\infty} \frac{t^5}{(1+t^3)^3}\,dt = \frac13\int_1^{\infty}\frac{u-1}{u^3}\,du = \frac13\int_1^{\infty}\left(\frac{1}{u^2}-\frac{1}{u^3}\right)du = \frac13\left[-\frac1u+\frac{1}{2u^2}\right]_1^{\infty} = \frac13\cdot\frac12 = \frac16\]

Step 5: Combine

\[A = 9a^2\left[\frac16 - 2\cdot\frac16\right] = 9a^2\left(-\frac16\right) = -\frac{3a^2}{2}\]

The negative sign reflects the loop's orientation as traversed by increasing $t$; taking the area as positive:

\[\boxed{A = \frac{3a^2}{2} \text{ square units}}\]

Step 6: Numerical check

For $a=1$: $A=1.5$ square units.

Summary

\[\boxed{A = \frac{3a^2}{2} \text{ square units}}\]

Key insight: The Folium of Descartes, studied by Descartes and Fermat, encloses a loop of area exactly $\frac{3a^2}{2}$; the curve also has the asymptote $x+y+a=0$.